On the unicity and location of orbits related to the non-trivial zeros of the zeta function: outlining a proof for the Riemann Hypothesis based on proba.

Intuition

The infinitely sectioned fish rod described above is bent by being pulled ortogonally downwards towards a given atractor.

Moments

Consider the series \(\zeta(s), Re(s)=2, Im(s)=0\). The corresponding segmented object’s length converges to:
\[\zeta(2)=\sum_{n= 1}^{\infty}\frac{1}{n^{2}}\]
\[=\frac{1}{1^{2}}+\frac{1}{2^{2}}+\frac{1}{3^{2}}+...\] \[\zeta(2)=\frac{\pi^{2}}{6}\]
The identity by Euler which solved the Basel problem.
We shall now consider our series with imaginary numbers:
\[\zeta(s)=\sum_{n= 1}^{\infty}\frac{1}{n^{s}}, Re(s)>1,Im(s)\neq 0\]

Consider if the imaginary parte of the previous series, \(\zeta(2)\), was \(Im(s)=\frac{\pi}{3}i\), then:
\[\zeta(2 + \frac{pi}{3})=\sum_{n= 1}^{\infty}\frac{1}{n^{2+\frac{\pi}{3}i}}\] \[=\sum_{n= 1}^{\infty}\frac{1}{n^{2}}*\frac{1}{n^{\frac{\pi}{3}i}}\]

By the definition, this is the dot product among vectors with lengths determined by the generating series \(\zeta(2)\) and \(\zeta(\frac{\pi}{3}i)\).

\[\zeta(2 + \frac{\pi}{3}) = \vec{\zeta(2)} \cdot \vec{\zeta(\frac{\pi}{3}i)} = \mid \zeta(2)\mid \mid\zeta(\frac{\pi}{3}i)\mid cos(\theta) \]

The value of \[\zeta(a+bi)\] converges to the dot product of vectors due to the sum \(a+bi\) in the exponent turning the expression into a product of perpendicular vectors:
\[\zeta(a+bi) = \sum_{1}^{\infty}\frac{1}{n^{a+bi}}\] \[=\sum_{1}^{\infty}\frac{1}{n^{a}}\frac{1}{n^{bi}}\] \(\vec{\zeta(a)}\), superposing the real axis and magnitude \(\sum_{1}^{\infty}\frac{1}{n^{a}}\), and perpendicular to \(\vec{\zeta(bi)}\), superposing the imaginary axis with magnitude \(\sum_{1}^{\infty}\frac{1}{n^{bi}}\).

Through vector addition, the next segment form and angle \(\theta\) with the continuation of the previous one. The fist segment crosses the origin with angle \(\theta\)

\[z = e^{\theta i} \rightarrow = cos(\theta) + sin(\theta)i\] \(z_{s_{n}}=z_{s_{n-1}}+z_{s_{n-2}}+...+z_{s_{1}}\)

The fisher rod analogy: A fish rod has several mathematical similarities with our problem. Consider the idea of an endeless fish rod, whose infinitesimal moving parts eventually become small and thin as a line. Its final size is the number asymptotically approaching by adding smaller pieces according to the series previously presented.

This peculiar fish rod has a spinner. When it is pulled towards the bottom, one may spin it and this will cause the rod to rotate periodically, perpendicular to the real axis.

Consider the orbit formed by the last point in each segment in a spiral towards the orbit, the attractor.

The vectorial decomposition of the spiral is \(z_{s_{n}}=z_{s_{n-1}}+z_{s_{n-2}}+...+z_{s_{1}}\).

When \(\zeta(s)\) approaches 0, that the vectorial sum approaches the center of the real axis. This is equal to rotate the object enough to cancel its projection on the real axis.

The first moment is \(M_{1}\). If the \(n_{th}\) line segment length is given by \(\frac{1}{n^{s}}\), we evaluate the moment around the real axis by the sum the distances \(M_{1} = \sum_{1}^{\infty}\frac{1}{n^{s}}\).

Each rod is uniquely determined by its momentums. The second momentum is: \(M_{2} = \sum_{1}^{\infty}(\frac{1}{n^{s}})^{2}\). Each \(Re(s)\) determines a family of rods. We hereby show that the scenario in which \(Re(s)=\frac{1}{2}\) is the one containing non-trivial zeta zeros.

Non-trivial zeros

Assume the line along \(Re(s)=\frac{1}{2}\) defines the atractor of an envelope defined by the orbits of \(\zeta\) in the critical strip \(0<s<1\) as a function of \(s\). It follows that this spiral is the one in which rotations from the vectorial decompostion cancel.

The second moment \(M_{2}\)(e.g: moment of inertia) of the spiral formed by \(\zeta(s)\) is then given by the generating series \[\zeta(2s)=\sum_{1}^{\infty}\frac{1}{n^2s}\].

The trivial orbits intersecting the real axis are the ones in which the second moment increases by an equal factor to its angle of rotation.

These are given by \(\zeta(-2ks), k=1,2,4,...\), as a direct consequence of
the trivial period for \(\sin{(\frac{\pi s}{2})}\) in the analytic continuation. \(\zeta{(s)} = 2^s \pi^{s-1} \sin{\left(\frac{\pi s}{2}\right)} \Gamma{(1-s)} \zeta{(1-s)}\)

Although this is only definable for the analytic continuation, the intuition holds for the series. We could also show that multiplying \(s\) by negative multiples of 2 is the same as: \[s=-(a+bi) \rightarrow 2s=2k*-(a+bi)\]. \[\sum_{1}{\infty}\frac{1}{n^2a}*\frac{1}{n^2bi}\]

For \(Re(s)>1\) it is easy to visualize that large dimensions of \(Re(s)\) result in orbits with large segments which do not rotate enough to turn the opposite way. The analytical continuation project those orbits symmetrically, emerging with associated angles in the left quadrants.

The operation changes the dot product by squaring both parts at each weigthed product of ortogonal parts. \[\vec{s'}^{2}=\vec{Re(s)}^2+\vec{Im(s)}^2\] Hence, the mapping \[f : \zeta(s) \rightarrow \zeta(2s), Re(s)<0, Im(s)=0\]. preserves the angle \(\theta\) with the center \(\zeta(-2s)\) when the geodesics of the system have no null curvature.

Riemann Hypothesis

The next (non-trivial) zeros are supposed to be related to the \(\zeta\) function when \(Re(s)=\frac{1}{2}\) and several seemenly related identities have been shown.

Excluding the trivial zeros, \(\zeta(s)=0\) only when \(0<s<1\). The envelope of spirals centered at the orbit \(\zeta(\frac{1}{2}+bi)\) is called critical strip.

The orbit \(\zeta(\frac{1}{2}+bi)\) is the vectorial decomposition of the orbit, or the sum of rod segments.
\[\sum_{1}^{\infty}\frac{1}{n^{(\frac{1}{2}+bi)}}\]

How to find the adequate orbit? In order for extra zeros to happen in the orbit, the net vectorial sum projection on the real axis must be 0.

Consider \(s\) along the interval \(S = 0 < 1/2+bi < 1\). No rotation, \(Im(s)=0\), is associated with the divergent series: \[\frac{1}{\sqrt 1}+\frac{1}{\sqrt 2}+\frac{1}{\sqrt 3}+\frac{1}{\sqrt 4}+\frac{1}{\sqrt 5}+\frac{1}{\sqrt 6}+...\] This is the sum of line segments with lenghts \(\sum_{1}{\infty}\frac{1}{sqrt{n}}\). And the analytic continuation of \(\zeta\), which would reflect the first moment related to the real axis: \[\zeta(1/2)=-(\sqrt{2}+1)\sqrt{\frac{2}{\pi}} \int_0^\infty\frac{\sqrt{x}\ \text{d}x}{\text{cosh}^2\ x}\] \[\sim-1.46035450880958\]

The second moment of the line segments is: \[M_{2}=\sum_{1}^{\infty}\frac{1}{{n^{\frac{1}{2}}}}^{2}\] \[=\sum_{1}{\infty}\frac{1}{n}\] Therefore, the \(M_{2}\) associated with 0 rotation is that of is the diverging harmonic series. Its truncation (harmonic number) can be expressed analytically as:

\[H = \gamma + \psi(n+1)\] where \(\gamma\) is the Euler-Mascheroni constant and \(\psi(x)\) is the logarithmic derivative of the gamma function. The digamma, \[\frac{d}{ds}ln{\Gamma(s)}\] \[=\frac{\Gamma(s)'}{\Gamma(s)}\]

\(\gamma\) is the pole of first order, its principal value exists and is equal to the Euler–Mascheroni constant, the limit of \(-ln (n) + \sum_{k=1}^{n}1/k\) as \(n\) increases to infinity. Also, \[\gamma = \int_{0}^{\infty} (-\frac{1}{x} + \frac{1}{floor(x)}) dx\] And \[\psi(s+1) = \psi(s)+\frac{1}{s}\] \[=\frac{\Gamma(s)'}{\Gamma(s)}+\frac{1}{s}\]

Notice that the moment of inertia \(M_{2}\) for large values of s given by: \[M_{2} \sim H = \gamma + \psi(s+1)\] Where \(\gamma\) is limit of an integral and \(\psi(s+1)=\psi(s)+\frac{1}{s}\) : \[= \int_{0}^{s} (-\frac{1}{s} + \frac{1}{floor(s)}) ds + \frac{d}{ds}ln{\Gamma(s)} + 1/s\] Taking the derivative of the second moment with respect to \(s\) \[\frac{d}{ds} ln(x) = 1/x, \frac{d}{dx} x^{-1} = \frac{-1}{x^2} \rightarrow\] \[= \frac{d}{ds}\int_{0}^{s}(-\frac{1}{s} + \frac{1}{floor(s)}) + \frac{d^2}{ds}ln{\Gamma(s)} + \frac{d}{ds}\frac{1}{s}\] The second derivative of the \(\Gamma\) function is the first derivative of \(\psi(s)\). Its Laurent series at \(s=0\) and \(\infty\): \[1/s^2+ \pi^2/6 + s\psi(s)''+\frac{\pi^2s^2}{30}+ O(s^3)\] \[1/s + 1/2s^2 + 1/6x^3 - 1/30s^5 + O((1/x)^6)\]

For \(s \sim 0\): \[M_{2} \sim (-\frac{1}{s} + \frac{1}{floor(s)}) + 1/s^2+ \pi^2/6 + s\psi(s)''+\frac{\pi^2s^2}{30}+ O(s^3) - 1/s^2\] \[= (-\frac{1}{s} + \frac{1}{floor(s)}) + \pi^2/6 + s\psi(s)''+\frac{\pi^2s^2}{30}+ O(s^3)\] For \(s \sim \infty\): \[M_{2} \sim (-\frac{1}{s} + \frac{1}{floor(s)}) + 1/s + 1/2s^2 + 1/6x^3 - 1/30s^5 + O((1/x)^6) - 1/s^2\] \[ = (-\frac{1}{s} + \frac{1}{floor(s)}) + 1/s + 1/6x^3 - 1/30s^5 + O((1/x)^6)\]

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\[= (-\frac{1}{s} + \frac{1}{floor(s)}) + \frac{d}{ds} \frac{1}{\Gamma(s)} - s^{-2}\]

\[= (-\frac{1}{s} + \frac{1}{floor(s)}) + \frac{d}{ds} \frac{1}{\Gamma(s)} - s^{-2}\]

\[= \int_{0}^{s} (-\frac{1}{s} + \frac{1}{floor(s)}) dx + \psi(s) + \frac{1}{s}\] The value is given by the logarithmic derivative of the gamma function until that moment \[psi(s)\] added with s-1\(\frac{1}{s}\)

Hence, the moment is given

The digamma is the derivative logarithmic of the gamma function. Considering products of successive rotations, the gamma function continues the factorials to real and complex numbers.

This shows that \[M_{2}\] can be approximated by a linear combination of a constant \((\gamma)\) and

Each value of \(s\) determine unique moments for the line segments. Considering the dot products,

For the atractor of an orbit to be centered at zero, the second moment totally compensate the rotation by vectorial sums. \[M_{2}(s)=\sum_{1}^{\infty} \frac{1}{{n^{\frac{1}{2}}}}^{2}\]

components must be co-periodic. The line segments should decrease their size
First moment The second moment must increase periodically at a spefic rate.

Consider the

\[\zeta(s)=\sum_{1}{\infty}\frac{1}{n^s}\] \[\zeta(2s)=\sum_{1}^{\infty}\frac{1}{n^2s}\].

\(\zeta(0) = \frac{-1}{2}\).

The non-trivial zeros. The \(Im(s)=\) Consider that a hypothetical fish rod is being pull.

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